搜索在多維數組的話，爪哇

-2

我想搜索在Java中的文字解謎的話。搜索，如所述的是在水平，垂直和對角線。我創建了一個數組，但我不知道如何創建一個字符串，並在我的字符串中搜索單詞。我需要知道如何獲得一個保留表格的所有值的字符串，以及如何能夠鍵入一個單詞並在此處進行搜索。搜索在多維數組的話，爪哇

我知道搜索的詞用的indexOf函數來完成，但我不知道如何執行它。 import java.util.Scanner;

public class main { 

    public static void main(String[] args) { 
     // TODO Auto-generated method stub 
     int IntegerPosition; 
     int IntegerPosition2; 
     String position=""; 
     String word=""; 
     Scanner s = new Scanner(System.in); 
     String content=""; 
     String[][] sopa = { 
       {"X","F","E","K","J","U","I","R","S","H"}, 
       {"Z","H","S","W","E","R","T","G","O","T"}, 
       {"B","R","A","B","F","B","P","M","V","U"}, 
       {"D","W","E","R","O","O","J","L","L","W"}, 
       {"U","T","O","N","I","R","O","B","C","R"}, 
       {"O","P","R","O","V","I","I","K","V","B"}, 
       {"N","I","Q","U","E","N","T","N","S","A"}, 
       {"O","V","U","L","R","O","S","S","O","T"}, 
       {"A","S","A","X","J","T","R","R","I","T"}, 
       {"R","K","M","E","P","U","B","O","T","A"} 
     }; 
     for (int i = 0; i < sopa[0].length; i++){ 
      for(int j = 0; j < sopa[i].length; j++){ 
       content += sopa[i][j]; 
      } 
      System.out.println(content); 
      content = ""; 
     } 
     System.out.println("Type the word you are looking for"); 
     word = s.next(); 

     for (int i = 0; i < sopa[0].length-1; i++){//t1.length 
      for(int j = 0; j < sopa[i].length-1; j++){ 


      } 
     } 
     System.out.println(content); 
     content = ""; 
    } 

}

來源

2016-01-26 FireFilip

表現出任何試圖將該自行解決，我們會盡量引導朝着正確的方向。 – Idos

非常好的參考：http://stackoverflow.com/questions/4795383/how-to-traverse-though-a-2d-char-array-searching-for-words-in-java – user2004685

首先，你應該聲明「找到一個詞」的含義。我想你想在每一行和每列中找到字母序列。那對角線呢？向後？環繞？

兩種解決方案浮現在腦海中：

使用字符串指數：
- 建設的所有字符的字符串。這需要針對每個方向（水平，垂直，對角線）完成，但只有在向前搜索的搜索項顛倒後才能執行。爲了有效實施StringBuilder是你的朋友。
- 使用String.indexOf找到索引術語的出現次數。最後，您必須從字符串位置計算行和列，並且如果不允許換行，請檢查該字是否跨越任何行/列邊界。
- 我會用這個，如果我不得不尋找很多條款。
使用陣列
- 還爲每個方向（水平，垂直，對角線）
- 查找您的陣列中的檢索詞的第一個字母的出現次數（通過簡單的迭代）。請注意，當該術語不適合行/列時，您可以停止，因此對於6個字母的單詞，可以跳過最後5行/列。
- 如果您發現錨點（即匹配字母），請檢查該詞的後續字母。取消不匹配，否則您找到了一個發生。

對於更成熟的配套實施，該Boyer-Moore algorithm可能會感興趣的。

來源

2016-01-26 15:27:27 penguineer

搜索在多維數組的話，爪哇

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